Problem: Solve for $x$ and $y$ using elimination. ${4x-3y = 28}$ ${5x+3y = 62}$
Answer: We can eliminate $y$ by adding the equations together when the $y$ coefficients have opposite signs. Add the equations together. Notice that the terms $-3y$ and $3y$ cancel out. $9x = 90$ $\dfrac{9x}{{9}} = \dfrac{90}{{9}}$ ${x = 10}$ Now that you know ${x = 10}$ , plug it back into $\thinspace {4x-3y = 28}\thinspace$ to find $y$ ${4}{(10)}{ - 3y = 28}$ $40-3y = 28$ $40{-40} - 3y = 28{-40}$ $-3y = -12$ $\dfrac{-3y}{{-3}} = \dfrac{-12}{{-3}}$ ${y = 4}$ You can also plug ${x = 10}$ into $\thinspace {5x+3y = 62}\thinspace$ and get the same answer for $y$ : ${5}{(10)}{ + 3y = 62}$ ${y = 4}$